Max files per directory in S3

amazon-s3directoryfilelimitperformance

If I had a million images, would it be better to store them in some folder/sub-folder hierarchy or just dump them all straight into a bucket (without any folders)?

Would dumping all the images into a hierarchy-less bucket slow down LIST operations?

Is there a significant overhead in creating folders and sub folders on the fly and setting up their ACLs (programatically speaking)?

Best Solution

S3 doesn't respect hierarchical namespaces. Each bucket simply contains a number of mappings from key to object (along with associated metadata, ACLs and so on).

Even though your object's key might contain a '/', S3 treats the path as a plain string and puts all objects in a flat namespace.

In my experience, LIST operations do take (linearly) longer as object count increases, but this is probably a symptom of the increased I/O required on the Amazon servers, and down the wire to your client.

However, lookup times do not seem to increase with object count - it's most probably some sort of O(1) hashtable implementation on their end - so having many objects in the same bucket should be just as performant as small buckets for normal usage (i.e. not LISTs).

As for the ACL, grants can be set on the bucket and on each individual object. As there is no hierarchy, they're your only two options. Obviously, setting as many bucket-wide grants will massively reduce your admin headaches if you have millions of files, but remember you can only grant permissions, not revoke them, so the bucket-wide grants should be the maximal subset of the ACL for all its contents.

I'd recommend splitting into separate buckets for:

totally different content - having separate buckets for images, sound and other data makes for a more sane architecture
significantly different ACLs - if you can have one bucket with each object receiving a specific ACL, or two buckets with different ACLs and no object-specific ACLs, take the two buckets.

Related Solutions

Bash – How to get the source directory of a Bash script from within the script itself

#!/usr/bin/env bash

SCRIPT_DIR="$( cd -- "$( dirname -- "${BASH_SOURCE[0]}" )" &> /dev/null && pwd )"

is a useful one-liner which will give you the full directory name of the script no matter where it is being called from.

It will work as long as the last component of the path used to find the script is not a symlink (directory links are OK). If you also want to resolve any links to the script itself, you need a multi-line solution:

#!/usr/bin/env bash

SOURCE="${BASH_SOURCE[0]}"
while [ -h "$SOURCE" ]; do # resolve $SOURCE until the file is no longer a symlink
  DIR="$( cd -P "$( dirname "$SOURCE" )" >/dev/null 2>&1 && pwd )"
  SOURCE="$(readlink "$SOURCE")"
  [[ $SOURCE != /* ]] && SOURCE="$DIR/$SOURCE" # if $SOURCE was a relative symlink, we need to resolve it relative to the path where the symlink file was located
done
DIR="$( cd -P "$( dirname "$SOURCE" )" >/dev/null 2>&1 && pwd )"

This last one will work with any combination of aliases, source, bash -c, symlinks, etc.

Beware: if you cd to a different directory before running this snippet, the result may be incorrect!

Also, watch out for $CDPATH gotchas, and stderr output side effects if the user has smartly overridden cd to redirect output to stderr instead (including escape sequences, such as when calling update_terminal_cwd >&2 on Mac). Adding >/dev/null 2>&1 at the end of your cd command will take care of both possibilities.

To understand how it works, try running this more verbose form:

#!/usr/bin/env bash

SOURCE="${BASH_SOURCE[0]}"
while [ -h "$SOURCE" ]; do # resolve $SOURCE until the file is no longer a symlink
  TARGET="$(readlink "$SOURCE")"
  if [[ $TARGET == /* ]]; then
    echo "SOURCE '$SOURCE' is an absolute symlink to '$TARGET'"
    SOURCE="$TARGET"
  else
    DIR="$( dirname "$SOURCE" )"
    echo "SOURCE '$SOURCE' is a relative symlink to '$TARGET' (relative to '$DIR')"
    SOURCE="$DIR/$TARGET" # if $SOURCE was a relative symlink, we need to resolve it relative to the path where the symlink file was located
  fi
done
echo "SOURCE is '$SOURCE'"
RDIR="$( dirname "$SOURCE" )"
DIR="$( cd -P "$( dirname "$SOURCE" )" >/dev/null 2>&1 && pwd )"
if [ "$DIR" != "$RDIR" ]; then
  echo "DIR '$RDIR' resolves to '$DIR'"
fi
echo "DIR is '$DIR'"

And it will print something like:

SOURCE './scriptdir.sh' is a relative symlink to 'sym2/scriptdir.sh' (relative to '.')
SOURCE is './sym2/scriptdir.sh'
DIR './sym2' resolves to '/home/ubuntu/dotfiles/fo fo/real/real1/real2'
DIR is '/home/ubuntu/dotfiles/fo fo/real/real1/real2'

Git – How to add a blank directory to a Git repository

Another way to make a directory stay (almost) empty (in the repository) is to create a .gitignore file inside that directory that contains these four lines:

# Ignore everything in this directory
*
# Except this file
!.gitignore

Then you don't have to get the order right the way that you have to do in m104's solution.

This also gives the benefit that files in that directory won't show up as "untracked" when you do a git status.

Making @GreenAsJade's comment persistent:

I think it's worth noting that this solution does precisely what the question asked for, but is not perhaps what many people looking at this question will have been looking for. This solution guarantees that the directory remains empty. It says "I truly never want files checked in here". As opposed to "I don't have any files to check in here, yet, but I need the directory here, files may be coming later".

Best Solution

Related Solutions

Bash – How to get the source directory of a Bash script from within the script itself

Git – How to add a blank directory to a Git repository

Related Question